Answer:
The correct answer is a
Explanation:
The electric field is given by the relation
F = q E
where The force is the Coulomb force and q is a positive test charge, therefore the electric field has the same direction as the electric force.
Consequently if the charge is positive the field must go out of the charge, if the charge is negative the electric field must be directed towards the charge.
Consequently, in this exercise we are told that the lines are directed towards the load, therefore the load must have a negative sign.
The correct answer is a
A car accelerates uniformly in a straight line with acceleration 10m/s 2 and travels 150m in a time interval of 5s. How far will it travel in the next 5s?
Explanation:
Given:
Acceleration of car = 10 m/s
Distance travelled in 5 sec = 150 m
To find:
Distance travelled in the next 5 seconds
Concept:
There are 2 ways to app this kind of questions .
Either , we can find the total distance travelled in 10 seconds and then subtract 150 m from it.
Whereas , you can find out the final velocity at the end of the 5th seconds. Using equation of motion, then get the distance travelled in next 5 seconds.
Calculation:
v² = u² + 2as
=> ( u + at)² = u² + 2as
=> u² + 2uat + a²t² = u² + 2as
=> 2uat + (at)² = 2as
=> 2u (10)(5) + (10 × 5)² = 2 (10)(150)
=> 100u = 3000 - 2500
=> 100u = 500
=> u = 5 m/s
Distance travelled in 10 seconds :
s = ut + ½at²
=> s = (5 × 10) + ½(10)(10)²
=> s = 50 + 500
=> s = 550 m
Distance travelled in the 2nd half will be :
d = 550 - 150
=> d = 400 m
So final answer is :
a bird of mass 2350g is flying at a height of 20.0m above the ground with a speed of 10m/s2 calculate it's potential energy
Explanation:
p.e =mgh
given: m=2350g=2.35kg h=20 g=9.8m/s
p.e=mgh
=2.35kg×20.0m×9.8
=460.6j
I am not sure
The late news reports the story of a shooting in the city. Investigators think that they have recovered the weapon and they run ballistics tests on the pistol at the firing range. If a 0.050-kg bullet were fired from the handgun with a speed of 400 m/s and it traveled 0.080 m into the target before coming to rest, what force did the bullet exert on the target?
Answer:
50000 N
Explanation:
From the question given above, the following data were obtained:
Mass (m) of bullet = 0.050 kg
velocity (v) = 400 m/s
Distance (s) = 0.080 m
Force (F) =?
Next, we shall determine the acceleration of the bullet. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Final velocity (v) = 400 m/s
Distance (s) = 0.080 m
Acceleration (a) =?
v² = u² + 2as
400² = 0 + (2 × a × 0.08)
160000 = 0 + 0.16a
160000 = 0.16a
Divide both side by 0.16
a = 160000 / 0.16
a = 1×10⁶ m/s²
Finally, we shall determine the force exerted by the bullet on the target. This can be obtained as follow:
Mass (m) of bullet = 0.050 kg
Acceleration (a) of bullet = 1×10⁶ m/s²
Force (F) =?
F = ma
F = 0.050 × 1×10⁶
F = 50000 N
Thus, the bullet exerted a force of 50000 N on the target.
What is 6 Fahrenheit in celcious
Answer:
6 Fahrenheit, converted over to Celsius, would be -14.444444
A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 39.0 cm/scm/s . What are:
a. The amplitude of the subsequent oscillations?
b. The block's speed at the point where x= 0.750 A?
Answer:
a
[tex]A = 0.081 \ m[/tex]
b
The value is [tex]u = 0.2569 \ m/s[/tex]
Explanation:
From the question we are told that
The mass is [tex]m = 0.750 \ kg[/tex]
The spring constant is [tex]k = 17.5 \ N/m[/tex]
The instantaneous speed is [tex]v = 39.0 \ cm/s= 0.39 \ m/s[/tex]
The position consider is x = 0.750A meters from equilibrium point
Generally from the law of energy conservation we have that
The kinetic energy induced by the hammer = The energy stored in the spring
So
[tex]\frac{1}{2} * m * v^2 = \frac{1}{2} * k * A^2[/tex]
Here a is the amplitude of the subsequent oscillations
=> [tex]A = \sqrt{\frac{m * v^ 2 }{ k} }[/tex]
=> [tex]A = \sqrt{\frac{0.750 * 0.39 ^ 2 }{17.5} }[/tex]
=> [tex]A = 0.081 \ m[/tex]
Generally from the law of energy conservation we have that
The kinetic energy by the hammer = The energy stored in the spring at the point considered + The kinetic energy at the considered point
[tex]\frac{1}{2} * m * v^2 = \frac{1}{2} * k x^2 + \frac{1}{2} * m * u^2[/tex]
=> [tex]\frac{1}{2} * 0.750 * 0.39^2 = \frac{1}{2} * 17.5* 0.750(0.081 )^2 + \frac{1}{2} * 0.750 * u^2[/tex]
=> [tex]u = 0.2569 \ m/s[/tex]
Is the range of the projectile dependent or independent of the projectile's mass? Explain.
a 150kg roller coaster SITTING ON THE TOP OF A 200M HILL HAS HOW MUCH POTETNTIAL ENERGY
Answer:
Epot = 294300 [J]
Explanation:
Potential energy is defined as the product of mass by height by gravitational acceleration. The height is measured with respect to the reference level. At this reference level the potential energy is equal to zero.
[tex]E_{pot}=m*g*h\\[/tex]
where:
m = mass = 150 [kg]
g = gravity acceleration [m/s²]
h = elevation = 200 [m]
[tex]E_{pot}=150*9.81*200\\E_{pot}=294300 [J][/tex]
Cork has a density of about 0.60 g/cm3 . Cork will partially float in water which has a density of 1.0 g/cm3 . If the piece of cork has a total volume of 5.0 cm3 what volume of cork is below the water
Answer:
3 cm³
Explanation:
Density of the cork = 0.6 g/cm³
Density of water = 1 g/cm³
Volume of cork = 5 cm³
We all know that the formula for density is given as
Density = mass/volume,
The mass of the cork is
Mass = density * volume
Mass = 0.6 * 5
Mass = 3 gram
Given that the density is 0.6, and it is partially floating, then we can say that the volume of the cork below the water is
5 * 0.6 = 3 cm³
The volume of cork is below the water 3 cm³.
Given data:
The density of Cork is, [tex]\rho = 0.60 \;\rm cm^{3}[/tex].
The density of water is, [tex]\rho_{w} = 1.0 \;\rm g/cm^{3}[/tex].
The total volume of piece of cork is, [tex]V = 5.0 \;\rm cm^{3}[/tex].
In the given problem, we can use the concept of density. The mass of substance occupied per unit volume is known as Density of substance. The expression for the density is given as,
Density = mass/volume,
And, the mass of the cork is
Mass = density * volume
Mass = 0.6 * 5
Mass = 3 gram
Given that the density is 0.6, and it is partially floating, then we can say that the volume of the cork below the water is,
= 5 * 0.6 = 3 cm³
Thus, we can conclude that the volume of cork is below the water 3 cm³.
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The acceleration of gravity is -9.8 m/s2. A sling shot fires a rock straight up into the air with a speed of +39.2 m/s. 1. what is its velocity after 2 seconds?
Given that,
The acceleration of gravity is -9.8 m/s²
Initial velocity, u = 39.2 m/s
Time, t = 2 s
To find,
The final velocity of the shot.
Solution,
Let v is the final velocity of sling shot. Using first equation of motion to find it.
v = u +at
Here, a = -g
v = u-gt
v = (39.2)-(9.8)(2)
v = 19.6 m/s
So, its velocity after 2 seconds is 19.6 m/s.
An astronaut weighing 190 lbs on Earth is on a mission to the Moon and Mars.
Required:
a. What would he weigh in newtons when he is on the Moon?
b. How much would he weigh in newtons when he is on Mars, where the acceleration due to gravity is 0.38 times that on Earth?
Answer:
The weight is defined as:
W = m*g
where:
m = mass
g = gravitational acceleration.
We know that in Earth the astronaut weights 190 lb-f (this is force, not mass, the correct unit here is 190 lb*m/s^2)
then:
190 lb*m/s^2 = m*9.8m/s^2
(190 lb*m/s^2)/(9.8m/s^2) = 19.39 lb
Now we know the mass of the astronaut.
a) wieght on the moon in Newtons.
Newtons uses kilograms as the units of mass, then we need to rewrite the mass of the astronaut in kg.
we know that 1lb = 0.454 kg
Then 19.39 lb is equal to: 19.39*0.454 kg = 8.8 kg
We know that the acceleration due to gravity on the Moon is one-sixth that on Earth.
then: g = (9.8m/s^2)/6
And the weight of the astronaut in the moon will be:
W = 8.8 kg*(9.8m/s^2)/6 = 14.37 N
b) The weight on mars, where the acceleration due to gravity is 0.38 times that on Earth, we have:
g = (9.8m/s^2)*0.38
then the weight will be:
W = 8.8kg*(9.8m/s^2)*0.38 = 32.77 N
A bag of shells weighs 1.5 N. What is its mass? approximately 1.5 kilograms approximately 1.5 pounds approximately 150 grams approximately 1.5 grams
Answer:
approximately 150 grams
Explanation:
The weight of an object is a product of its mass and acceleration of gravity acting upon it. Since weight (W) is force, it is measured in Newtons (N) or Kgms-²
W = mg
Where;
W = weight (N)
m = mass (kg)
g = acceleration due to gravity
g is a constant, which is approximately 10m/s²
Therefore, according to this question, the mass of a bag of shells that weighs 1.5N can be calculated thus:
m = W/g
m = 1.5/10
m = 0.15 kg
Converting 0.15kg to grams, we multiply by 1000 i.e 0.15 × 1000 = 150 grams.
Hence, the mass of the object is approximately 150grams.
i)x-1 by 4=3
ii)3x+6 by 2=3 by 2
iii)3x-1by5=2x+3by7
Answer:
i) x = 13
ii) x = -1
iii) x = 2
Explanation:
Given the following expressions, we are to find the corresponding value of x.
1) x-1 by 4=3
x-1/4=3
Cross multiply
x-1 = 4*3
x-1 = 12
x = 12+1
x = 13
2) Given 3x+6 by 2 = 3 by 2
3x+6/2=3/2
Cross multiply
2(3x+6) = 2(3)
2(3x) + 2(6) = 6
6x + 12 = 6
6x = 6-12
6x = -6
x = -6/6
x = -1
3) 3x-1by5=2x+3by7
This can also be written as:
(3x-1)/5 = (2x+3)/7
Cross multiply
7(3x-1) = 5(2x+3)
Open the bracket
7(3x) - 7(1) = 5(2x) + 5(3)
21x - 7 = 10x + 15
Collect like terms
21x - 10x = 15 + 7
11x = 22
x = 22/11
x = 2
If the Earth were flat, then the shadows of two towers at two different places on the Earth would:__________.
A. be different lengths with the one further South being longer
B. be different lengths with the one further North being longer
C. be different lengths with the one further East being longer
D. be the exact same length
An object increases its velocity from 22 m/s to 36 m/s in 5 s. What is the acceleration of the
object?
Add process please
Explanation:
Using Kinematics,
we have a = (v - u) / t.
Therefore a = (36m/s - 22m/s) / 5s = 2.8m/s².
The figure above shows the net force exerted on an object as a function of the position of the object. The object starts from rest at position x = 0 m and acquires a speed of 3.0 m / s after traveling the distance of 0.090 m shown above. What is the mass of the object?
Answer:
0.06 Kg
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Final velocity (v) = 3.0 m/s
Distance (s) = 0.09 m
Net Force (F) = 3 N
Mass (m) =?
Next, we shall determine the acceleration of the object. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Final velocity (v) = 3.0 m/s
Distance (s) = 0.09 m
Acceleration (a) =?
v² = u² + 2as
3² = 0² + (2 × a × 0.09)
9 = 0 + 0.18a
9 = 0.18a
Divide both side by 0.18
a = 9 / 0.18
a = 50 m/s²
Finally, we shall determine the mass of the object. This can be obtained as follow:
Net Force (F) = 3 N
Acceleration (a) = 50 N
Mass (m) =?
F = ma
3 = m × 50
Divide both side by 50
m = 3 / 50
m = 0.06 Kg
Therefore, the mass of the object is 0.06 Kg
The required value for the mass of object is 0.06 kg.
Given data:
The speed of object is, v = 3.0 m/s.
The distance travelled by object is, s = 0.090 m.
From the given graph in the question, we obtain the following details as,
The net force applied on the object is, [tex]F_{net}=3\;\rm N[/tex].
The initial velocity is, u = 0 m/s.
Now using the third kinematic equation of motion to obtain the acceleration of object as,
v² = u² + 2as
3² = 0² + (2 × a × 0.09)
9 = 0 + 0.18a
9 = 0.18a
Solving as,
a = 9 / 0.18
a = 50 m/s²
Now, as per the Newton's second law of motion, we have the expression for the applied force as,
[tex]F_{net}= ma[/tex]
Here, m is the mass of object.
Solving as,
[tex]3= m \times 50\\\\m = 0.06 \;\rm kg[/tex]
Thus, we can conclude that the required value for the mass of object is 0.06 kg.
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a boulder with a mass of about 1.5 x 10^5 kg falls and strikes the ground at 70 m/s how much kinetic energy dies the boulder deliver to the ground PLEASE HELP NEED HELP ASAP
Explanation:
K.E=1/2mv^2
m=1.5×10^5
v=70
K.E=1/2×1.5×70
K.E=52.5×10^5
A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is the magnitude of the current in the other wire
Answer:
The current is [tex]I_b = 400 \ A[/tex]
Explanation:
From the question we are told that
The length of the segment is [tex]l = 2.50 \ m[/tex]
The current is [tex]I_a = 1000 \ A[/tex]
The force felt is [tex]F = 4.0 \ N[/tex]
The distance of the second wire is [tex]d = 5.0 \ cm = 0.05 \ m[/tex]
Generally the current on the second wire is mathematically represented as
[tex]I_b = \frac{2 \pi * r * F }{ l * \mu_o * I_a }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4 \pi * 10^{-7} \ N/A^2[/tex]
=> [tex]I_b = \frac{2 * 3.142 * 0.05 * 4 }{ 2.50 * 4\pi *10^{-7} * 1000 }[/tex]
=> [tex]I_b = 400 \ A[/tex]
20 POINTS
An object released from rest at time t = 0 slides down a frictionless
incline a distance of 2 m during the time interval from t=0 s to t = 1 s.
The distance traveled by the object during the time interval from t = 2s
to t = 3s is: (A) 15 m (B) 10 m (C) 5 m (D) 2 m (E) 1m
Answer:
10m
Explanation:
let's take the acceleration as a constant throughout the complete motion...
therefore first let's find the acceleration
[tex]s = ut + \frac{1}{2} a {t}^{2} \\ 2 = \frac{1}{2} a {1}^{2} \\ a = 4m {s}^{ - 2} [/tex]
then we have to find v1
apply V = u + at
v = 4×1
= 4ms^-1
lets find the distance travel by the object DURING THE TIME INTERVAL 1-2
[tex]s = ut + \frac{1}{2} a {t}^{2} \\ s = 4 \times 1 + \frac{1}{2} \times 4 \times {1}^{2} \\ s= 6m [/tex]
then let's find the V2
[tex] {v}^{2} = {u}^{2} + 2as \\ {v}^{2} = {4 }^{2} + 2 \times 4 \times 6 \\ {v}^{2} = 64 \\ v = \sqrt{64} = 8m {s}^{ - 1} [/tex]
then let's find the distance travel by the object during time interval 2s to 3s[tex]s = ut + \frac{1}{2} a {t}^{2} \\ s = 8 \times 1 + \frac{1}{2} \times 4 \times {1}^{2} \\ s = 10m [/tex]
The distance traveled by the object during the time interval from t = 2 seconds to t = 3 seconds would be 10 meters, therefore the correct answer is option B.
What are the three equations of motion?There are three equations of motion given by Newton
v = u + at
S = ut + 1/2×a×t²
v² - u² = 2×a×s
As given in the problem, an object released from rest at time t = 0 slides down a frictionless incline a distance of 2 m during the time interval from t=0 seconds to t = 1 seconds.
2 = ut + 1/2*a*t²
2 = 0 + 0.5×a×1²
a = 2 / 0.5
a = 4 meters / second²
Now to find the velocity after the 2 seconds,
v = u + at
v = 0 + 4×2
v = 8 m/s
Now by using the second equation of motion,
S = ut + 1/2×a×t²
S = 8×1 + 0.5×4×1²
S = 8 + 2
S = 10 meters
Thus, The distance traveled by the object during the time interval from t = 2 seconds to t = 3 seconds would be 10 meters, therefore the correct answer is option B.
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An elastic conducting material is stretched into a circular loop of 11.2 cm radius. It is placed with its plane perpendicular to a uniform 0.880 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 68.8 cm/s. What emf is induced in volts in the loop at that instant?
Answer:
0.426 volts
Explanation:
It is given that,
The radius of a circular loop, r = 11.2 cm = 0.112 m
An elastic conducting material is stretched into a circular loop.
It is placed with its plane perpendicular to a uniform 0.880 T magnetic field.
The radius of the loop starts to shrink at an instantaneous rate of 68.8 cm/s, dr/dt = 0.688 m/s
We need to find the emf induced in the loop at that instant.
[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\=\dfrac{d}{dt}(BA)\\\\=\dfrac{d}{dt}(\pi r^2 B)\\\\=\pi B\dfrac{d}{dt}(r^2)\\\\=2\pi B r\dfrac{dr}{dt}\\\\=2\pi \times 0.88\times 0.112\times 0.688\\\\=0.426\ V[/tex]
So, the magnitude of induced emf is 0.426 volts.
Two prominent research groups came to the same surprising conclusion after taking measurements of the luminosity of Type Ia supernovae at great distances, this being that the universe is accelerating while it expands.
A. True
B. False
Answer:
A. True
Explanation:
Type la supernova can be described as a phenomenon whereby two stars are in orbit with one another .
Recently , two prominent research groups came to the same surprising conclusion after taking measurements of the luminosity of Type Ia supernovae at great distances that the universe is accelerating while it expands.
A raindrop charged to 10 μC experiences electric field of 10,000 V/m from a tip of a tree branch. What is the electric force acting on the raindrop? (1μC= 10^-6C)
a. 0.001 N
b. 100 N
c. 0.1N
d. 0.9 N
Answer:
(C). The magnitude of electric force acting on the raindrop is 0.1 N.
Explanation:
Given;
charge of the raindrop, Q = 10 μC = 10 x 10⁻⁶ C
electric field strength, E = 10,000 V/m
The electric force acting on the raindrop is given as;
F = EQ
where;
F is the electric force
Substitute the given values and solve for F.
F = (10,000)(10 x 10⁻⁶)
F = 0.1 N
Therefore, the magnitude of electric force acting on the raindrop is 0.1 N.
A small object carrying a charge of -4.00 nC is acted upon by a downward force of 24.0 nN when placed at a certain point in an electric field. What are the magnitude and direction of the electric field at the point in question
Answer:
[tex]E = -6 \ N/C[/tex]
Generally given that the electric field is negative it mean that its direction is opposite to that of the force
Explanation:
From the question we are told that
The charge on the small object is [tex]Q = -4.00 \ nC = -4.00 *10^{-9} \ C[/tex]
The force is [tex]F = 24 \ nN = 24 *10^{-9} \ N[/tex]
Generally the magnitude of the electric field is mathematically represented as
[tex]E = \frac{F}{Q}[/tex]
=> [tex]E = \frac{ 24 *10^{-9}} {-4 *10^{-9 }}[/tex]
=> [tex]E = -6 \ N/C[/tex]
Generally given that the electric field is negative it mean that its direction is opposite to that of the force
The displacement of a car is a function of time as follows: x(t)=25+3.0t², with x is in meters. Find the average velocity between t1 = 1.0s and t2 = 4.0s.
Answer: 15m/s
Explanation: Average Velocity is vector describing the total displacement of an object and the time taken to change its position. It is represented as:
[tex]v=\frac{\Delta x}{\Delta t}[/tex]
At t₁ = 1.0s, displacement x₁ is:
[tex]x(1)=25+3(1)^{2}[/tex]
x(1) = 28
At t₂ = 4.0s:
[tex]x(4)=25+3(4)^{2}[/tex]
x(4) = 73
Then, average speed is
[tex]v=\frac{73-28}{4-1}[/tex]
v = 15
The average velocity of a car between t₁ = 1s and t₂ = 4s is 15m/s
A box is pulled 6 meters across the ground at a constant velocity by a horizontally applied force of 50 newtons. At the same time, kinetic friction acts on the box as it slides so that it eventually stops when the pulling force stops. What is the magnitude of the frictional force
Answer:
The magnitude of the frictional force is 50 newtons.
Explanation:
The frictional force can be found as follows:
[tex] \Sigma F = ma [/tex]
[tex] F - F_{\mu} = ma [/tex]
Where:
F: is the applied force = 50 N
[tex]F_{\mu}[/tex]: is the frictional force
m: is the box's mass
a: is the acceleration
Since the box is pulled at a constant velocity, a = 0, so:
[tex] F - F_{\mu} = 0 [/tex]
[tex] F = F_{\mu} = 50 N [/tex]
Therefore, the magnitude of the frictional force is equal to the applied force, that is to say, 50 newtons.
I hope it helps you!
g a 100 m3 container with 1/3 water with pressure of 100 MPa drops to 90 Mpa how much heat transfer is required to bring back to the initial condition
Answer:
Q = 3.33 108 J
Explanation:
This is an exercise in thermodynamics, specifically isobaric work
W = V ([tex]P_{f}[/tex]- P₀)
They tell us that we have ⅓ of the volume of the container filled with water
V = ⅓ 100
V = 33.3 m³
let's calculate
W = 33.3 (90-100) 10⁶
W = - 3.33 10⁸ J
To bring the system to its initial condition if we use the first law of thermodynamics
ΔE = Q + W
as we return to the initial condition the change of internal energy (ΔE) is zero
W = -Q
therefore the required heat is
Q = 3.33 108 J
What acceleration will you give a 22.4 kg box if you push it with a force of 83.1N
Answer:
mass =22.4kg
force=83.1N
a=?
f=ma
a=f/m
a=83.1/22.4
a=3.70m/s^2
A Light spiral spring is loaded with
a mass of 50g and it extends by
10cm.
(is calculate the period of
small vertical oscillations.
Answer:
0.63 s
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 50 g
Extention (e) = 10 cm
Period (T) =?
Next, we obtained 50 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
50 g = 50 g × 1 Kg / 1000 g
50 g = 0.05 kg
Next, we shall convert 10 cm to m. This is illustrated below:
100 cm = 1 m
Therefore,
10 cm = 10 cm × 1 m / 100 cm
10 cm = 0.1 m
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass = 0.05 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = mg
F = 0.05 × 9.8
F = 0.49 N
Next, we shall determine the spring constant of the spring.
Extention (e) = 0.1 m
Force (F) = 0.49 N
Spring constant (K) =?
F = Ke
0.49 = K × 0.1
Divide both side by 0.1
K = 0.49 /0.1
K = 4.9 N/m
Finally, we shall determine the period as follow:
Mass = 0.05 Kg
Spring constant (K) = 4.9 N/m
Pi (π) = 3.14
Period (T) =?
T = 2π√(m/k)
T = 2 × 3.14 × √(0.05 / 4.9)
T = 6.28 × √(0.05 / 4.9)
T = 0.63 s
Thus, the period of oscillation is 0.63 s
Which of the following objects will have more kinetic energy?
Kinetic Energy =
(Joules)
12 x mass x (velocity)?
(kg) (m/s)
KE = 12 mv
O A 6 kg ball thrown a 8 m/s.
O A 2 kg ball thrown at 15 m/s.
O A 4 kg ball thrown at 10 m/s.
Determination of the energy imparted to matter by radiation is the subject of ______. A. photometry B. magnification C. dosimetry D. optometry
Answer:
dosimetry
Explanation:
dosimetry is the determination of energy imparted to matter by radiation
If a nucleus was as big as a nonpareil, an atom would be ____
Answer:
small marble
Explanation:
A nonpareil are confectionery tiny ball items that are made up of starch and sugar. It is very small of the size of sugar crystal or sand grains. They were the miniature version of the comfits. They are generally opaque white but bow available in all colors.
In the context, if the nucleus is compared to the size of a nonpareil then its atom would be of the size of small size marble. An atom is bigger in size than that of nucleus as the nucleus is located inside the atoms.